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Tatiana [17]
3 years ago
5

if a voltage is applied to a capacitor, current flows easily at first and then slows as the capacitor becomes charged. Inductors

behave just the opposite, in that they reluctantly pass current when a voltage is first applied, and then the current passes easily as time passes. If the input voltage is suddenly raised from zero to some constant value, sketch the current in the capacitor, iC, and the inductor, iL, as a function of time. What is the steady-state current in the capacitor and inductor?

Engineering
1 answer:
DENIUS [597]3 years ago
7 0

Answer: photo!

Explanation:

For the capacitor, the current as a function of time will be

I(t)=V/R e^(-t/RC)

In steady-state, time tends to infinite, so current tends to zero.

For the inductor, the current as a function of time is

I(t)=V/R (1-e^(-tR/L))

In steady-state, time tends to infinite, so current tends to V/R

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In the case above,  poor connection at the pressure cycling switch  and also a faulty A/C clutch coil could be the cause.

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a. 9947 m

b. 99476 times

c. 2*10^11 molecules

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a) To find the mean free path of the air molecules you use the following formula:

\lambda=\frac{RT}{\sqrt{2}\pi d^2N_AP}

R: ideal gas constant = 8.3144 Pam^3/mol K

P: pressure = 1.5*10^{-6} Pa

T: temperature = 300K

N_A: Avogadros' constant = 2.022*10^{23}molecules/mol

d: diameter of the particle = 0.25nm=0.25*10^-9m

By replacing all these values you obtain:

\lambda=\frac{(8.3144 Pa m^3/mol K)(300K)}{\sqrt{2}\pi (0.25*10^{-9}m)^2(6.02*10^{23})(1.5*10^{-6}Pa)}=9947.62m

b) If we assume that the molecule, at the average, is at the center of the chamber, the times the molecule will collide is:

n_{collision}=\frac{9947.62m}{0.05m}\approx198952\  times

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3 years ago
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